Concerns about climate change and CO2 reduction have initiated the commercial production of blends of biodiesel (e.g., from renewable sources) and petrodiesel (from fossil fuel). Random blended fuel samples of size 35 are tested in a lab to ascertain the bio/total carbon ratio (X). If the true (i.e. population) mean is 0.948 and the true (i.e. population) standard deviation is 0.006, what is the distribution of Xbar?

Answer:

a) [tex]df= k-1[/tex]

b) [tex]df= (k_1-1)(k_2-1)[/tex]

Step-by-step explanation:

Previous concepts

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Solution to the problem

Let's define some notation:

N= Total number of individuals for a population

n = Total of individuals selected from a sample

n1= number of objects/individuals with characteristic 1

n2= number of objects/individuals with characteristic 2

k1= number of levels for the variable or factor 1

k2= number of levels for the variable or factor 2

Part a

For a chi square goodness of fit test the degrees of freedom are given by:

[tex]df= k-1[/tex]

Where k represent the total number of categories on the godness of fit test.

Part b

For a chi square test of independence the degrees of freedom are given by:

[tex]df= (k_1-1)(k_2-1)[/tex]

Where k1= number of levels for the variable or factor 1, k2= number of levels for the variable or factor 2 for the chi square test for independence.

In statistical analysis, degrees of freedom (df) are calculated differently based on the type of test. For a chi-square goodness-of-fit test, df equals the number of categories minus one. For a chi-square test for independence, df equals the product of subtracting one from the number of columns and rows.

The degrees of freedom (df) play an important role in many statistical tests, including the chi-square tests for goodness-of-fit and independence. These degrees of freedom are calculated differently based on the type of test. In a chi-square goodness-of-fit test, the degrees of freedom is calculated as df = k - 1, where k represents the number of categories or groups in the dataset.

On the other hand, for a chi-square test for independence, the degrees of freedom are calculated using the formula df = (number of columns - 1)(number of rows - 1), where columns and rows represent the different categories or groups present in a contingency table used in the test of independence. In this context, the null hypothesis for this test states that the two factors being considered are independent.

It's important to remember that degrees of freedom often represent the quantity of the 'free' or varying elements in your dataset and directly impact the test results and conclusions. So, having an accurate calculation is crucial in your statistical analysis.

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Answer: Reject the eight- ounces claim.

Step-by-step explanation:

For left tailed test , On a normal curve the rejection area lies on the left side of the critical value.

It means that if the observed z-value is less than the critical value then it will fall into the rejection region other wise not.

As per given ,

Objective : A coffee-dispensing machine is supposed to deliver eight ounces of liquid or less.

Then ,

[tex]H_0: \mu=8\\\\H_a: \mu<8[/tex] , since alternative hypothesis is left-tailed thus the test is an left-tailed test.

the critical value for z for a one-tailed test with the tail in the left end is -1.645 and the obtained value is -1.87.

Clearly , -1.87 < -1.645

⇒ -1.87 falls under rejection region.

⇒  Decision : Reject null hypothesis.

i.e. we reject the eight- ounces claim.

Answer:

[tex]t=\frac{39-40}{\frac{8.8}{\sqrt{70}}}=-0.9508[/tex]      

[tex]p_v =P(t_{69}<-0.9508)=0.1725[/tex]  

If we compare the p value and the significance level given for example [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we FAIL to reject the null hypothesis, and the the actual mean is not significantly less than 40 months.

Step-by-step explanation:

1) Data given and notation      

[tex]\bar X=39[/tex] represent the sample mean

[tex]s=8.8[/tex] represent the standard deviation for the sample      

[tex]n=70[/tex] sample size      

[tex]\mu_o =40[/tex] represent the value that we want to test    

[tex]\alpha[/tex] represent the significance level for the hypothesis test.    

t would represent the statistic (variable of interest)      

[tex]p_v[/tex] represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.      

We need to conduct a hypothesis in order to determine if the mean  is less than 40 months, the system of hypothesis would be:      

Null hypothesis:[tex]\mu \geq 40[/tex]      

Alternative hypothesis:[tex]\mu < 40[/tex]      

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:      

[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)      

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic      

We can replace in formula (1) the info given like this:      

[tex]t=\frac{39-40}{\frac{8.8}{\sqrt{70}}}=-0.9508[/tex]      

Calculate the P-value      

First we need to calculate the degrees of freedom given by:  

[tex]df=n-1=70-1=69[/tex]  

Since is a one-side lower test the p value would be:      

[tex]p_v =P(t_{69}<-0.9508)=0.1725[/tex]  

Conclusion      

If we compare the p value and the significance level given for example [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we FAIL to reject the null hypothesis, and the the actual mean is not significantly less than 40 months.      

Answer:

Step-by-step explanation:

For a field trip, the school bought 39 sandwiches and 42 bags of chips. Each sandwich cost 4.35. That means the cost of 39 sandwiches would be

39 × 4.35 = 169.65

Each bag of chip costs 15 cents. Converting 15 cents to dollars, it becomes

15/100 =$ 0.15

Therefore, the cost of 42 bags of chips would be

42 × 0.15 = 6.3

Total cost of 39 sandwiches and 42 bags of chips would be

169.65 + 6.3 = 175.95

Answer:

1. [tex]t_{0.01/2,(10)}=3.169[/tex]

2. Confidence interval = [15.2, 25.2]

Step-by-step explanation:

X = 20.2

S = 5.19

n = 11

Step 1 of 2: Critical value

α = 1 - 0.99 = 0.01

df = 11 - 1 = 10

Looking at t distribution table with α = 0.01 and df = 10, we find

[tex]t_{0.01/2,(10)}=3.169[/tex]

Step 2 of 2: 99% confidence interval

[tex]std-err=\frac{S}{\sqrt{n}}=\frac{5.19}{\sqrt{11}}=1.5648[/tex]

[tex]X+t_{0.01/2,10}*std-err[/tex]

20.2 + 3.169*1.5648

20.2 + 4.9595

25.2

[tex]X-t_{0.01/2,10}*std-err[/tex]

20.2 - 3.169*1.5648

20.2 - 4.9595

15.2

Confidence interval = [15.2, 25.2]

Hope this helps!

Answer:

Part (a): The mean of the distribution is 2.

Part (b): The standard deviation of the distribution is 1.525

Step-by-step explanation:

Consider the provided information.

Part (a) What is the mean of the distribution?

The mean of the sampling distribution of [tex](\bar x_1 -\bar x_2)[/tex] is (µ₁ − µ₂).

It is given that μ₁ = 31, σ₁ = 3, μ₂ = 29, and σ₂ = 2.

Mean of distribution = (x₁ - x₂)

Mean of distribution = (31 - 29) = 2

Hence, the mean of the distribution is 2.

Part (b) What is the standard deviation of the distribution?

If the two samples are independent, the standard deviation of the sampling distribution is: [tex]\sigma_{(x_1-x_2)}=\sqrt{\left(\dfrac{\sigma^2_1}{n_1}+\dfrac{\sigma^2_2}{n_2}\right)}}[/tex]

Substitute the respective values in the above formula.

[tex]\sigma_{(x_1-x_2)}=\sqrt{\left(\dfrac{3^2}{4}+\dfrac{2^2}{53}\right)}}[/tex]

[tex]\sigma_{(x_1-x_2)}=\sqrt{\left(\dfrac{9}{4}+\dfrac{4}{53}\right)}}[/tex]

[tex]\sigma_{(x_1-x_2)}\approx1.525[/tex]

Hence, the standard deviation of the distribution is 1.525

The approximate sampling distribution of x1 - x2 has a mean equal to μ1 - μ2 and a standard deviation equal to √((σ1^2/n1) + (σ2^2/n2)).

A sampling distribution is the probability distribution of a statistic, like the mean or proportion, calculated from multiple samples of the same size drawn from a population. It provides insights into the variability of the statistic and is fundamental in statistical inference.

The approximate sampling distribution of x1 - x2 has a mean equal to the difference of the population means, μ1 - μ2. The standard deviation of the distribution is equal to the square root of the sum of the variances of the two populations divided by their respective sample sizes, √((σ1^2/n1) + (σ2^2/n2)).

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Answer:

[tex]z=\frac{7.25-7.5}{\frac{1.4}{\sqrt{10}}}=-0.565[/tex]  

[tex]p_v =P(Z<-0.565)=0.286[/tex]  

a) If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true average is not significantly less than 7.5.  

b) [tex]\chi^2 =\frac{10-1}{1.96} 4 =18.367[/tex]  

[tex]p_v =P(\chi^2 >18.367)=0.0311[/tex]

If we compare the p value and the significance level provided we see that [tex]p_v >\alpha[/tex] so on this case we have enough evidence in order to FAIL reject the null hypothesis at the significance level provided. And that means that the population variance is not significantly higher than 1.96.

Step-by-step explanation:

Assuming this info: "Suppose birth weights follow a normal distribution with mean 7.5 pounds and standard deviation 1.4 pounds"

1) Data given and notation  

[tex]\bar X=7.25[/tex] represent the sample mean  

[tex]s=1.2[/tex] represent the sample standard deviation

[tex]\sigma=1.4[/tex] represent the population standard deviation

[tex]n=10[/tex] sample size  

[tex]\mu_o =7.5[/tex] represent the value that we want to test  

[tex]\alpha=0.05,0.01[/tex] represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

[tex]p_v[/tex] represent the p value for the test (variable of interest)  

2) State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is less than 7.5, the system of hypothesis would be:  

Null hypothesis:[tex]\mu \geq 7.5[/tex]  

Alternative hypothesis:[tex]\mu < 7.5[/tex]  

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

[tex]z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}[/tex] (1)  

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

3) Calculate the statistic  

We can replace in formula (1) the info given like this:  

[tex]z=\frac{7.25-7.5}{\frac{1.4}{\sqrt{10}}}=-0.565[/tex]  

4)P-value  

Since is a left tailed test the p value would be:  

[tex]p_v =P(Z<-0.565)=0.286[/tex]  

5) Conclusion  

Part a

If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true average is not significantly less than 7.5.  

Part b

A chi-square test is "used to test if the variance of a population is equal to a specified value. This test can be either a two-sided test or a one-sided test. The two-sided version tests against the alternative that the true variance is either less than or greater than the specified value"

[tex]n=10[/tex] represent the sample size

[tex]\alpha=0.01[/tex] represent the confidence level  

[tex]s^2 =4 [/tex] represent the sample variance obtained

[tex]\sigma^2_0 =1.96[/tex] represent the value that we want to test

Null and alternative hypothesis

On this case we want to check if the population variance increase, so the system of hypothesis would be:

Null Hypothesis: [tex]\sigma^2 \leq 1.96[/tex]

Alternative hypothesis: [tex]\sigma^2 >1.96[/tex]

Calculate the statistic  

For this test we can use the following statistic:

[tex]\chi^2 =\frac{n-1}{\sigma^2_0} s^2[/tex]

And this statistic is distributed chi square with n-1 degrees of freedom. We have eveything to replace.

[tex]\chi^2 =\frac{10-1}{1.96} 4 =18.367[/tex]

Calculate the p value

In order to calculate the p value we need to have in count the degrees of freedom , on this case 9. And since is a right tailed test the p value would be given by:

[tex]p_v =P(\chi^2 >18.367)=0.0311[/tex]

In order to find the p value we can use the following code in excel:

"=1-CHISQ.DIST(18.367,9,TRUE)"

Conclusion

If we compare the p value and the significance level provided we see that [tex]p_v >\alpha[/tex] so on this case we have enough evidence in order to FAIL reject the null hypothesis at the significance level provided. And that means that the population variance is not significantly higher than 1.96.

Answer:

A .55

Step-by-step explanation:

Given that Leroy and Fred play chess at a club every Wednesday. The probability that Leroy will lose is .3, that he will stalemate is .5, and that he will win is .2. The probability that Fred will lose is .25, that he will stalemate is .4, and that he will win is .35.

lose stale win

Leroy 0.3 0.5 0.2

Fred 0.25 0.4 0.35

Probability that atleast one wins = Prob Leroy wins + Prob fred wins - prob both wins

(using addition theorem on probability)

But prob of both winning is 0

Hence required prob = 0.2 + 0.35 = 0.55

Option a is right

Answer:

0.48

Step-by-step explanation:

Answer:

a) The 99% confidence interval would be given by (589.588;731.038)

b) If we see the confidence interval the 620 is included on the interval we don't have enough evidence to reject the rating is 620. But since we need a score that at least 620 and our lower limit is 589.588 we cant conclude that all the clients would have a score that at least of 620.  

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The data is:

661 595 548 730 791 678 672 491 492 583 762 624 769 729 734 706

Part a

Compute the sample mean and sample standard deviation.  

In order to calculate the mean and the sample deviation we need to have on mind the following formulas:  

[tex]\bar X= \sum_{i=1}^n \frac{x_i}{n}[/tex]  

[tex]s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}[/tex]  

=AVERAGE(661, 595, 548, 730, 791, 678, 672, 491 ,492, 583, 762 ,624 ,769, 729, 734, 706)

On this case the average is [tex]\bar X= 660.313[/tex]

=STDEV.S(661, 595, 548, 730, 791, 678, 672, 491 ,492, 583, 762 ,624 ,769, 729, 734, 706)

The sample standard deviation obtained was s=95.898

Find the critical value t* Use the formula for a CI to find upper and lower endpoints

In order to find the critical value we need to take in count that our sample size n =16<30 and on this case we don't know about the population standard deviation, so on this case we need to use the t distribution. Since our interval is at 99% of confidence, our significance level would be given by [tex]\alpha=1-0.99=0.01[/tex] and [tex]\alpha/2 =0.005[/tex]. The degrees of freedom are given by:

[tex]df=n-1=16-1=15[/tex]

We can find the critical values in excel using the following formulas:

"=T.INV(0.005,15)" for [tex]t_{\alpha/2}=-2.95[/tex]

"=T.INV(1-0.005,15)" for [tex]t_{1-\alpha/2}=2.95[/tex]

The confidence interval for the mean is given by the following formula:

[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]  

And if we find the limits we got:

[tex]660.313- 2.95\frac{95.898}{\sqrt{16}}=589.588[/tex]  

[tex]660.313+ 2.95\frac{95.898}{\sqrt{16}}=731.038/tex]  

So the 99% confidence interval would be given by (589.588;731.038)

Part b

If we see the confidence interval the 620 is included on the interval we don't have enough evidence to reject the rating is 620. But since we need a score that at least 620 and our lower limit is 589.588 we cant conclude that all the clients would have a score that at least of 620.  

Answer:

a) If we see the alternative hypothesis we see that we are conducting a bilateral test or two tail.

b) (-1.9886, 1.9886)

c) [tex]t=\frac{(100 -98.4)-(0)}{\sqrt{\frac{3.4^2}{41}}+\frac{5.6^2}{45}}=1.617[/tex]

d) [tex]p_v =2*P(t_{84}>1.617) =0.1096[/tex]

So with the p value obtained and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 is NOT significantly different than the mean for the group 2.  

e) [tex]p_v =2*P(t_{84}>1.617) =0.1096[/tex]

We can use the following excel code to calculate it: "=2*(1-T.DIST(1.617;84;TRUE))"

Step-by-step explanation:

The system of hypothesis on this case are:

Null hypothesis: [tex]\mu_1 = \mu_2[/tex]

Alternative hypothesis: [tex]\mu_1 \neq \mu_2[/tex]

Or equivalently:

Null hypothesis: [tex]\mu_1 - \mu_2 = 0[/tex]

Alternative hypothesis: [tex]\mu_1 -\mu_2\neq 0[/tex]

Our notation on this case :

[tex]n_1 =41[/tex] represent the sample size for group 1

[tex]n_2 =45[/tex] represent the sample size for group 2

[tex]\bar X_1 =100[/tex] represent the sample mean for the group 1

[tex]\bar X_2 =98.4[/tex] represent the sample mean for the group 2

[tex]s_1=3.4[/tex] represent the sample standard deviation for group 1

[tex]s_2=5.6[/tex] represent the sample standard deviation for group 2

Part a

If we see the alternative hypothesis we see that we are conducting a bilateral test or two tail.

Part b

On this case since the significance level is 0.05 and we are conducting a bilateral test we have two critical values, and we need on each tail of the distribution [tex]\alpha/2 = 0.025[/tex] of the area.

The distribution on this cas since we don't know the population deviation for both samples is the t distribution with [tex]df=n_1+n_2 -2= 41+45-2=84[/tex] degrees of freedom.

We can use the following excel codes in order to find the critical values:

"=T.INV(0.025,84)", "=T.INV(1-0.025,84)"

And we got: (-1.9886, 1.9886)

Part c

The statistic is given by this formula:

[tex]t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{\sqrt{\frac{s^2_1}{n_1}}+\frac{S^2_2}{n_2}}[/tex]

And now we can calculate the statistic:

[tex]t=\frac{(100 -98.4)-(0)}{\sqrt{\frac{3.4^2}{41}}+\frac{5.6^2}{45}}=1.617[/tex]

The degrees of freedom are given by:

[tex]df=41+45-2=84[/tex]

Part d

And now we can calculate the p value using the altenative hypothesis:

[tex]p_v =2*P(t_{84}>1.617) =0.1096[/tex]

So with the p value obtained and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 is NOT significantly different than the mean for the group 2.  

Part e

[tex]p_v =2*P(t_{84}>1.617) =0.1096[/tex]

We can use the following excel code to calculate it: "=2*(1-T.DIST(1.617;84;TRUE))"

Answer: B

Step-by-step explanation:

Answer:

the real answer is2 or d

Answer:

4.57ft  by 1.57 ft

Step-by-step explanation:

We are given that

Emma's square patio has been area=31 sq.ft

One dimension decrease by 1 foot and other dimension decrease by 4 feet.

We have to find the new dimensions of Emma's patio.

Let x be the side of Emma's square patio

We know that

Area of square=[tex]x^2[/tex]

[tex]x^2=31[/tex]

[tex]x=\sqrt{31}=5.57[/tex] ft

One dimension=[tex]5.57-1=4.57[/tex] ft

other dimension=[tex]5.57-4=1.57 ft [/tex]

Hence, the new dimension of Emma's patio is given by

4.57ft  by 1.57 ft

Answer: 505

Step-by-step explanation:

The formula to find the sample size n , if the prior estimate of the population proportion (p) is known:

[tex]n= p(1-p)(\dfrac{z}{E})^2[/tex] , where E=  margin of error and z = Critical z-value.

Let p be the population proportion of crashes.

Prior sample size = 250

No. of people experience computer crashes = 75

Prior proportion  of crashes [tex]p=\dfrac{75}{250}=0.3[/tex]

E= 0.04

From z-table , the z-value corresponding to 95% confidence interval = z=1.96

Required sample size will be :

[tex]n=0.3(1-0.3)(\dfrac{1.96}{0.04})^2[/tex] (Substitute all the values in the above formula)

[tex]n= (0.21)(49)^2= 0.21\times2401[/tex]

[tex]n= 504.21\approx505[/tex]   (Rounded to the next integer.)

∴ Required sample size = 505

Answer:

There should be at most 24 lucky numbers in the third bag.

Step-by-step explanation:

Initially, there are 200 numbers. Two bags with 100 each. There are 31+18 = 49 lucky numbers. So there is a 49/200 = 0.245 probability that a randomly selected number from a random bag is the lucky number.

Now with 300 numbers, we want this probability to be lower than 24.5%. So we should solve the following rule of three:

200 - 49

300 - x

[tex]200x = 300*49[/tex]

[tex]x = 1.5*49[/tex]

[tex]x = 73.5[/tex]

With the third bag, the probability will be the same if 73.5-49 = 24.5 lucky numbers are added. So there should be at most 24 lucky numbers in the third bag.

Answer:

She should add 30 ounces of the 12% vinegar she should add to 40 ounces of the 5% vinegar to obtain the needed vinegar with an 8% acidity level.

Step-by-step explanation:

Let X be ounces of the 12% vinegar she should add to 40 ounces of the 5% vinegar to obtain the needed vinegar with an 8% acidity level.

We have the equation

[tex]\frac{(X*0.12)+(40*0.05}{X+40} =0.08[/tex] when 8% acidity level is reached.

Solving the equation we have

X*0.12+2=0.08(X+40)=0.08X+3.2

That is 0.12X-0.08X=3.2-2=1.2

Finally 0.04X=1.2 and

X=30

Answer:

140

Step-by-step explanation:

To construct a subset of S with said property, we have two choices, include 3 in the subset or include four in the subset. These events are mutually exclusive because 3 and 4 can not both be elements of the subset.

First, let's count the number of subsets that contain the element 3.

Any of such subsets has five elements, but since 3 is already an element, we only have to select four elements to complete it. The four elements must be different from 3 and 4 (3 cannot be selected twice and the condition does not allow to select 4), so there are eight elements to select from. The number of ways of doing this is [tex]{}_8C_4=70[/tex].

Now, let's count the number of subsets that contain the element 4.

4 is already an element thus we have to select other four elements . The four elements must be different from 3 and 4 (4 cannot be selected twice and the condition does not allow to select 3), so there are eight elements to select from, so this can be done in [tex]{}_8C_4=70[/tex] ways.

We conclude that there are 70+70=140 required subsets of S.

Answer:

[tex]p_v =2*P(t_{(8)}<-1.195)=0.266[/tex]

c. should not be rejected

Step-by-step explanation:

1) Data given and notation

The data given is:

Method 1 : 7,5,6,7,5

Method 2: 5,9,8,7,6

[tex]\bar X_{1}=6[/tex] represent the mean for the method 1

[tex]\bar X_{2}=7[/tex] represent the mean for the method 2

[tex]s_{1}=1[/tex] represent the sample standard deviation for the method 1

[tex]s_{2}=1.58[/tex] represent the sample standard deviation for the method 2

[tex]n_{1}=5[/tex] sample size selected for the Consultant A

[tex]n_{2}=5[/tex] sample size selected for the Consultant B

[tex]\alpha=0.05[/tex] represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

[tex]p_v[/tex] represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean for the difference is equal to 0, the system of hypothesis would be:

Null hypothesis:[tex]\mu_{1}- \mu_{2}=0[/tex]

Alternative hypothesis:[tex]\mu_{1} - \mu_{2}\neq 0[/tex]

If we analyze the size for the samples both are less than 30 so for this case is better apply a t test to compare means, and the statistic is given by:

[tex]t=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}}[/tex] (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".

Calculate the statistic

We can replace in formula (1) the info given like this:

[tex]t=\frac{6-7}{\sqrt{\frac{1^2}{5}+\frac{1.58^2}{5}}}=-1.195[/tex] (1)

P-value

The first step is calculate the degrees of freedom, on this case:

[tex]df=n_{1}+n_{2}-2=5+5-2=8[/tex]

Since is a two sided test the p value would be:

[tex]p_v =2*P(t_{(8)}<-1.195)=0.266[/tex]

Conclusion

If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis. So the best option would be:

c. should not be rejected

Answer:  a. 0.8

Step-by-step explanation:

When we test whether the variances of the two populations are equal we use F- test.

Test statistic : [tex]F=\dfrac{s_1^2}{s_2^2}[/tex]

, where [tex]s_1^2[/tex]= sample variance from population 1.

[tex]s_2^2[/tex]= sample variance from population 2.

As per given , we have

[tex]s_1^2=8[/tex] and  [tex]s_2^2=10[/tex]

Then,  If we test whether the variances of the two populations are equal, the test statistic will be [tex]F=\dfrac{8}{10}=0.8[/tex]

Hence, the test statistic will have a value of 0.8 .

Thus , the correct answer is  a. 0.8 .

The 95% confidence interval for the population proportion of college students who work to pay for tuition is (0.4228, 0.5172) and the 99% confidence interval is (0.4086, 0.5314).

To calculate a confidence interval for a population proportion, we use the formula p ± Z*(√((p(1-p))/n)), where p is the sample proportion, Z is the Z-score associated with our desired confidence level, and n is the sample size.

In this case, p = 0.47 (47%), n = 450, and the Z-scores are 1.96 for a 95% confidence interval and 2.58 for a 99% confidence interval.

For the 95% confidence interval, we calculate:

0.47 ± 1.96*(√((0.47*(1-0.47))/450)) = 0.47 ± 0.0472, so the 95% confidence interval is (0.4228, 0.5172).

For the 99% confidence interval, we calculate:

0.47 ± 2.58*(√((0.47*(1-0.47))/450)) = 0.47 ± 0.0614, so the 99% confidence interval is (0.4086, 0.5314).

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Answer:

y = 8 is the equation of tangent.

Step-by-step explanation:

The equation of the tangent to the circle at (-6,8) is of the form:

y = mx + c

where m is the slope of the tangent and c is the y-intercept.

The point (-6,8) lies on the circle and the tangent line as well.

Hence (-6,8) satisfies the line equation:

8 = m(-6) + c ⇒ c-6m = 8 -------------1

We know that slope of two perpendicular lines are related as:

[tex]m_{1}\times m_{2}=-1[/tex]

At any point on the circle, the normal line at a point is always perpendicular to the tangent line at that point.

Hence :

[tex]m_{normal} \times m_{tangent}=-1[/tex]

We can find the slope of the normal at point (-6,8) as it passes through the centre of the circle (-6,4) by using the two-points formula for slope.

[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]

         [tex]=\frac{8-4}{-6+6}[/tex]

          = ∞

Slope of the normal is infinity and hence slope of tangent is -1/∞ = 0

Hence m=0

Putting m=0 in equation 1 we get:

c = 8

The equation of tangent line at (-6,8) is:

y = 8

Answer:

y = 8

Step-by-step explanation:

The equation of the tangent to the circle at (-6,8) is of the form:

y = mx + c

where m is the slope of the tangent and c is the y-intercept.

The point (-6,8) lies on the circle and the tangent line as well.

Hence (-6,8) satisfies the line equation:

8 = m(-6) + c ⇒ c-6m = 8 -------------1

Answer:

[tex]p_v = P(\chi^2_{29}>42.963)=1-0.954=0.0459[/tex]

b. between .025 and .05

Step-by-step explanation:

Previous concepts and notation

The chi-square test is used to check if the standard deviation of a population is equal to a specified value. We can conduct the test "two-sided test or a one-sided test".

[tex]\bar X [/tex] represent the sample mean

n = 30 sample size

s= 0.2 represent the sample deviation

[tex]\sigma_o =\sqrt{0.027}=0.164[/tex] the value that we want to test

[tex]p_v [/tex] represent the p value for the test

t represent the statistic

[tex]\alpha=[/tex] significance level

State the null and alternative hypothesis

On this case we want to check if the population standard deviation is less than 0.027, so the system of hypothesis are:

H0: [tex]\sigma \leq 0.027[/tex]

H1: [tex]\sigma >0.027[/tex]

In order to check the hypothesis we need to calculate the statistic given by the following formula:

[tex] t=(n-1) [\frac{s}{\sigma_o}]^2 [/tex]

This statistic have a Chi Square distribution distribution with n-1 degrees of freedom.

What is the value of your test statistic?

Now we have everything to replace into the formula for the statistic and we got:

[tex] t=(30-1) [\frac{0.2}{0.164}]^2 =42.963[/tex]

What is the approximate p-value of the test?

The degrees of freedom are given by:

[tex] df=n-1= 30-1=29[/tex]

For this case since we have a right tailed test the p value is given by:

[tex]p_v = P(\chi^2_{29}>42.963)=1-0.954=0.0459[/tex]

And the best option would be:

b. between .025 and .05

Answer:

Step-by-step explanation:

Given that a curve in polar coordinates is given by:

r=9+3cosθ

a) At point P, we have

[tex]θ=\frac{21\pi}{18}[/tex]

Substitute to get

[tex]r=9+cos \frac{21\pi}{18}\\=9.3932[/tex]

b) Cartesian coordinate is

[tex]x= rcos \theta=3.6934\\y =r sin \tjeta =8.6365[/tex]

c) At the origin r =0

when r =0

we have

[tex]9+3cos\theta=0\\cos\theta =-3[/tex]

Since cos cannot take values as -3 it doe snot pass through origin.

In summary, the polar coordinate r for point P is approximately 9.59. The Cartesian coordinates of P are approximately (0.99, -9.43). The given curve passes through the origin twice in the interval 0<θ<2π.

This question is about a curve in polar coordinates. The curve is given by r=9+3cosθ. Let's explore the three sub-questions in order:

a. Find polar coordinate r for P

Because Point P is at θ=21π/18, we can plug this into the equation r=9+3cosθ. This gives r = 9 + 3cos(21π/18) = 9.59.

b. Find Cartesian coordinates for point P.

To convert from polar to Cartesian coordinates, we use the equations x = rcosθ and y = rsinθ. Plugging in our values, we get x = 9.59cos(21π/18) and y = 9.59sin(21π/18). Calculating these gives us x approximately equal to 0.99 and y approximately equal to -9.43. So the Cartesian coordinates for point P are (0.99, -9.43).

c. How many times does the curve pass through the origin when 0<θ<2π?

The curve reaches the origin when r is 0. From the curve equation r = 9 + 3cosθ = 0, we can see the curve crosses the origin twice, at θ = 2π/3 and θ = 4π/3.

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Answer:

a) There is a 13% probability that a student has taken 2 or more semesters of Calculus.

b) 45% probability that a student has taken some calculus.

c) 87% probability that a student has taken no more than one semester of calculus.

Step-by-step explanation:

We have these following probabilities:

A 55% that a student hast never taken a Calculus course.

A 32% probability that a student has taken one semester of a Calculus course.

A 100-(55+32) = 13% probability that a student has taken 2 or more semesters of Calculus.

a) two or more semesters of Calculus?

There is a 13% probability that a student has taken 2 or more semesters of Calculus.

b) some Calculus?

At least one semester.

So there is a 32+13 = 45% probability that a student has taken some calculus.

c) no more than one semester of Calculus?

At most one semester.

So 55+32 = 87% probability that a student has taken no more than one semester of calculus.

The correct probabilities for the given scenarios are:

a) The probability that the first groupmate you meet has studied two or more semesters of Calculus is [tex]$\boxed{\frac{13}{100}}$[/tex].

 b) The probability that the first groupmate you meet has studied some Calculus (which includes those who have taken only one semester and those who have taken two or more semesters) is [tex]$\boxed{\frac{45}{100}}$[/tex].

c) The probability that the first groupmate you meet has studied no more than one semester of Calculus is [tex]$\boxed{\frac{32}{100}}$[/tex].

To calculate these probabilities, we use the information provided by the professor:

- 55% of the students have never taken a Calculus course.

- 32% have taken only one semester of Calculus.

- The rest, which is 100% - (55% + 32%) = 13%, have taken two or more semesters of Calculus.

Now, let's calculate the probabilities for each part of the question:

a) The probability that the first groupmate has studied two or more semesters of Calculus is the percentage of students who have taken two or more semesters, which is 13%. In probability terms, this is [tex]$\frac{13}{100}$[/tex]

b) The probability that the first groupmate has studied some Calculus is the sum of the percentages of students who have taken at least one semester of Calculus. This includes both those who have taken only one semester (32%) and those who have taken two or more semesters (13%). Therefore, the probability is 32% + 13% = 45%. In probability terms, this is [tex]$\frac{45}{100}$[/tex].

c) The probability that the first groupmate has studied no more than one semester of Calculus is the sum of the percentages of students who have never taken Calculus and those who have taken only one semester. This is 55% + 32% = 87%. However, since we are looking for the probability of no more than one semester, we exclude those who have taken two or more semesters, so we subtract the 13% from the 87%, giving us 87% - 13% = 74%. But this calculation is incorrect because we have double-counted the 32% of students who have taken only one semester. The correct calculation is to consider only the students who have never taken Calculus (55%) and those who have taken only one semester (32%), which gives us 55% + 32% - 0% (since we do not need to subtract any percentage this time) = 87%. In probability terms, this is [tex]$\frac{32}{100}$[/tex], as we are only considering those who have taken no Calculus or just one semester.

Answer:

B.  the matched pairs t test.

Step-by-step explanation:

A paired t-test is used to compare two population means where you have two samples in  which observations in one sample can be paired with observations in the other sample. For example  if we have Before-and-after observations (This problem) we can use it.  

Let put some notation  

x=first test value , y = second test value

x: 920, 830, 960, 910

y: 1010, 800, 1000, 980

The system of hypothesis for this case are:

Null hypothesis: [tex]\mu_y- \mu_x \leq 0[/tex]

Alternative hypothesis: [tex]\mu_y -\mu_x >0[/tex]

The first step is calculate the difference [tex]d_i=y_i-x_i[/tex] and we obtain this:

d: 90, -30, 40, 70

The second step is calculate the mean difference  

[tex]\bar d= \frac{\sum_{i=1}^n d_i}{n}= \frac{170}{4}=42.5[/tex]

The third step would be calculate the standard deviation for the differences, and we got:

[tex]s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =52.520[/tex]

The 4 step is calculate the statistic given by :

[tex]t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{42.5 -0}{\frac{52.520}{\sqrt{4}}}=1.618[/tex]

The next step is calculate the degrees of freedom given by:

[tex]df=n-1=4-1=3[/tex]

Now we can calculate the p value, since we have a right tailed test the p value is given by:

[tex]p_v =P(t_{(3)}>1.618) =0.1024[/tex]

So the p value is higher than any significance level assumed (0.05), so then we can conclude that we reject the null hypothesis. So we can conclude that the difference between the after and the initial score it's not significantly higher at 5% of significance.

Answer:

The cost function is [tex]C(x)=x^4-3x^2+5x+550[/tex].

Step-by-step explanation:

It is given that the marginal cost of manufacturing an item when x thousand items are produced is

[tex]\frac{dC}{dx}=4x^3-6x+5[/tex]

We need to find the cost function.

Multiply both sides by dx.

[tex]dC=(4x^3-6x+5)dx[/tex]

Integrate both sides to find the cost function.

[tex]\int dC=\int (4x^3-6x+5)dx[/tex]

[tex]\int dC=4\int x^3dx-6\int xdx+5\int 1 dx[/tex]

[tex]C(x)=4(\frac{x^4}{4})-6(\frac{x^2}{2})+5x+C[/tex]

where, C(x) is const function and C is a constant.

[tex]C(x)=x^4-3x^2+5x+C[/tex]

It is given that C(0)=550. Substitute x=0 in the above function.

[tex]C(0)=(0)^4-3(0)^2+5(0)+C[/tex]

[tex]C(0)=C[/tex]

[tex]550=C[/tex]

Therefore, the cost function is [tex]C(x)=x^4-3x^2+5x+550[/tex].

Answer:

The mean number favoring the substation is 12 citizens.

Step-by-step explanation:

For each citizen surveyed, there are only two possible outcomes. EIther they favored building a police substation in their neighborhood, or they opposed. This means that we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, with p probability, and X can only have two outcomes.

Has an expected value of:

[tex]E(X) = np[/tex]

In this problem, we have that:

[tex]n = 15, p = 0.8[/tex]

E(X) = 15*0.8 = 12.

The mean number favoring the substation is 12 citizens.

Final answer:

The mean number of citizens out of 15 who favor the construction of a police substation is 12, calculated by multiplying the sample size (15) by the probability of favoring the substation (0.80).

Explanation:

To find the mean number favoring the substation, we'll use the probability of a citizen favoring the substation to calculate the expected value in a sample of 15 citizens. Since 80% of the community favored building a police substation, the mean number favoring the substation can be computed by multiplying the sample size by the probability of favoring the substation.

Mean = Sample Size × Probability of Favoring the Substation

Mean = 15 × 0.80

Mean = 12

Therefore, the mean number of citizens out of 15 who favor the construction of the police substation is 12.

To fill out the normal distribution curve, use the average weight and standard deviation. To find the percentage of people who would receive a bag with a weight greater than a certain amount, calculate the Z-score and use a Z-table.

To fill out the normal distribution curve for this situation, we can use the given information. The bags have an average weight of 1,040 grams with a standard deviation of 25 grams. This means that the mean of the distribution is 1,040 and the standard deviation is 25. We can plot the normal distribution curve using these values.

To find the percentage of people who would receive a bag that had a weight greater than 1,115 grams, we need to find the area under the normal distribution curve to the right of 1,115. We can calculate this using a Z-score calculator or a Z-table. The Z-score for 1,115 is (1,115 - 1,040) / 25 = 3. From the Z-table, we can find that the area to the right of Z-score 3 is approximately 0.0013. This means that approximately 0.13% of people would receive a bag that weighs more than 1,115 grams.

Answer:

95% Confidence interval for the mean

[tex]142.8 \leq\mu\leq158.4[/tex]

Step-by-step explanation:

We have to calculate a 95% confidence interval for the mean of a finite population.

The error is multiplied by the following finite population correction factor:

[tex]cf=\sqrt{\frac{N-n}{N-1} }[/tex]

The standard deviation can be estimated as

[tex]\sigma=\frac{s}{\sqrt{n}} \sqrt{\frac{N-n}{N-1} } =\frac{24.4}{\sqrt{32} }* \sqrt{\frac{200-32}{200-1} }=3.963[/tex]

The 95% confidence interval has a z value of 1.96, so it becomes:

[tex]M-z*\sigma_c\leq\mu\leq M+z*\sigma_c\\\\150.6-1.96*3.963\leq\mu\leq 150.6+1.96*3.963\\\\ 142.8 \leq\mu\leq 158.4[/tex]

Answer:

lower limit =0.261

upper limit =0.369

Step-by-step explanation:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

Description in words of the parameter p

[tex]p[/tex] represent the real population proportion of people that have experienced bullying

X= 63 people in the random sample that have experienced bullying

n=200 is the sample size required  

[tex]\hat p=\frac{63}{200}=0.315[/tex] represent the estimated proportion of people that have experienced bullying

[tex]z_{\alpha/2}[/tex] represent the critical value for the margin of error  

The population proportion have the following distribution  

[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]

Numerical estimate for p

In order to estimate a proportion we use this formula:

[tex]\hat p =\frac{X}{n}[/tex] where X represent the number of people with a characteristic and n the total sample size selected.

[tex]\hat p=\frac{63}{200}=0.315[/tex] represent the estimated proportion of people that they were planning to pursue a graduate degree

Confidence interval

The confidence interval for a proportion is given by this formula  

[tex]\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]  

For the 90% confidence interval the value of [tex]\alpha=1-0.90=0.1[/tex] and [tex]\alpha/2=0.05[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.  

[tex]z_{\alpha/2}=1.64[/tex]  

And replacing into the confidence interval formula we got:  

[tex]0.315 - 1.64 \sqrt{\frac{0.315(1-0.315)}{200}}=0.261[/tex]  

[tex]0.315 - 1.64 \sqrt{\frac{0.315(1-0.315)}{200}}=0.369[/tex]  

And the 90% confidence interval would be given (0.261;0.369).  

We are confident at 90% that the true proportion of people that they were planning to pursue a graduate degree is between (0.261;0.369).  

lower limit =0.261

upper limit =0.369

The lower limit of the confidence interval is 0.270, and the upper limit is 0.360.

To find the 90% z-confidence interval for the true proportion of students who have experienced bullying, we can use the formula:

CI = p ± z * sqrt((p(1-p))/n)

Where:

p is the sample proportion (63/200 = 0.315)

z is the z-value for a 90% confidence interval (z = 1.645)

n is the sample size (200)

Plugging in these values, we get:

CI = 0.315 ± 1.645 * sqrt((0.315(1-0.315))/200)

Simplifying the equation gives us:

CI = 0.315 ± 0.045

So, the lower limit of the confidence interval is 0.315 - 0.045 = 0.270, and the upper limit is 0.315 + 0.045 = 0.360.

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Answer:

a. 8.66 minutes

Step-by-step explanation:

Since the flight times are uniformly distributed, the standard deviation can be calculated as follows:

[tex]\sigma = \frac{b-a}{\sqrt{12}}[/tex]

Where 'b' is the maximum flight time (150 minutes) and 'a' is the minimum flight time (120 minutes):

[tex]\sigma = \frac{150-120}{\sqrt{12}}\\\sigma = 8.66\ minutes[/tex]

The distribution's standard deviation is 8.66 minutes.